Question 801572


if you have {{{x-y=-3}}}, means you have an equation of a line

you can solve it for {{{y}}}

{{{x-y=-3}}}.....add {{{y}}} to both sides

{{{x-y+y=y-3}}}


{{{x-cross(y)+cross(y)=y-3}}}


{{{x =y-3}}}....add {{{3}}} to both sides

{{{x+3 =y-3+3}}}

{{{x+3 =y-cross(3)+cross(3)}}}

{{{x+3 =y }}} 

or {{{y=x+3 }}}.....this is equation of a line in slope-intercept form {{{y=mx+b}}} where {{{m=slope}}} and {{{b=y-intercept}}}

in your case {{{m=1}}} and {{{b=3}}}

so, your line has positive slope, means goes through I and III quadrant 

it also crosses y-axis at point ({{{0}}},{{{3}}})

we need one more point to graph it:

let's find {{{x}}} intercept if {{{y=0}}}

{{{0=x+3 }}}
{{{-3=x  }}}

so, {{{x}}} intercept is at ({{{-3}}},{{{0}}})


see it on a graph:


{{{ graph( 600,600, -10, 10, -10, 10, x+3) }}}